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Math Help - Finding the equation for a plane

  1. #1
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    Finding the equation for a plane

    The plane that contains the points (8,3,1), (2,6,3), and (4,6,2) has an equation of the form ax+by+cz=d. Find coefficients fr this equation, trying two different approaches to the problem. One method uses vectors, another does not.
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  2. #2
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    Since the points P_{1}, \;\ P_{2}, \;\ P_{3} lie in the plane, the vectors \overline{P_{1}P_{2}}=(-6,3,2) and
    \overline{P_{1}P_{3}}=(-4,3,1) are parallel to the plane.

    The cross product:

    P_{1}P_{2}\times{P_{1}P_{3}}=\begin{vmatrix}i&j&k\  \-6&3&-2\\-4&3&1\end{vmatrix}=-3i-2j-6k
    is normal to the plane.

    Use this normal and the point (8,3,1):

    -3(x-8)-2(y-3)-6(z-1)=0

    And we get:

    -3x-2y-6z+36=0

    \boxed{3x+2y+6z=36}

    Now, try the other method.
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